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Coulombs law

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Coulombs law

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Coulombs law

  1. 1. Electric Charge and Coulomb’s Law
  2. 2. Fundamental Charge: The charge on one electron. e = 1.6 x 10 -19 C Unit of charge is a Coulomb (C)
  3. 3. Two types of charge: Positive Charge: A shortage of electrons. Negative Charge: An excess of electrons. Conservation of charge – The net charge of a closed system remains constant.
  4. 4. + n + + + + + n n n n n - - - - - - Neutral Atom Number of electrons = Number of protons Nucleus Negative Atom Number of electrons > Number of protons -2e = -3.2 x 10-19 C - - Positive Atom Number of electrons < Number of protons +2e = +3.2 x 10-19 C
  5. 5. Electric Forces Like Charges - Repel Unlike Charges - Attract - +F F + + FF
  6. 6. Coulomb’s Law – Gives the electric force between two point charges. 2 21 r qq kF = k = Coulomb’s Constant = 9.0x109 Nm2 /C2 q1 = charge on mass 1 q2 = charge on mass 2 r = the distance between the two charges The electric force is much stronger than the gravitational force. Inverse Square Law
  7. 7. 2 21 r qq kF = If r is doubled then F is : If q1 is doubled then F is : If q1 and q2 are doubled and r is halved then F is : ¼ of F 2F 16F Two charges are separated by a distance r and have a force F on each other. q1 q2 r F F Example 1
  8. 8. Example 2 Two 40 gram masses each with a charge of 3μC are placed 50cm apart. Compare the gravitational force between the two masses to the electric force between the two masses. (Ignore the force of the earth on the two masses) 3μC 40g 50cm 3μC 40g
  9. 9. 2 21 r mm GFg = 2 11 )5.0( )04)(.04(. 1067.6 − ×= N13 1027.4 − ×≈ 2 21 r qq kFE = 2 66 9 )5.0( )103)(103( 100.9 −− ×× ×= N324.0≈ The electric force is much greater than the gravitational force
  10. 10. 5μC - 5μC Three charged objects are placed as shown. Find the net force on the object with the charge of -4μC. - 4μC F2 F1 F1 and F2 must be added together as vectors. 20cm 20cm cm282020 22 ≈+ 45º 45º 2 21 r qq kF = NF 5.4 )20.0( )104)(105( 109 2 66 9 1 = ×× ×= −− NF 30.2 )28.0( )104)(105( 109 2 66 9 2 = ×× ×= −− Example 3
  11. 11. F1 F2 45º 2.3cos45≈1.6 2.3sin45≈1.6 F1 = < - 4.5 , 0.0 > F2 = < 1.6 , - 1.6 >+ Fnet = < - 2.9 , - 1.6 > NFnet 31.36.19.2 22 ≈+= - 2.9 - 1.6 3.31 θ  29 9.2 6.1 tan 1 ≈      − − = − θ 3.31N at 209º 29º
  12. 12. Example 4 Two 8 gram, equally charged balls are suspended on earth as shown in the diagram below. Find the charge on each ball. qq 20º L = 30cmL = 30cm FEFE r =2(30sin10º)=10.4cm 2 2 2 21 r q k r qq kFE == 10º10º 30sin10º r
  13. 13. Draw a force diagram for one charge and treat as an equilibrium problem. FE Fg = .08N T q Tsin80º NT T 081. 80sin 08. 08.80sin ≈= =   Tcos80º Cq k q q k TFE 7 22 2 2 103.1 )104(. 014. 80cos)081(. 104. 80cos − ×= = = =  80º

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